Obtaining [latex]K_{a}[/latex] from [latex]K_{b}[/latex] or [latex]K_{b}[/latex] from [latex]K_{a}[/latex] Use Tables 17.1 and 17.2 to obtain the following at [latex]25^{\circ} \mathrm{C}[/latex] : a. [latex]K_{b}[/latex] for [latex]\mathrm{CN}^{-}[/latex]; b. [latex]K_{a}[/latex] for [latex]\mathrm{NH}_{4}^{+}[/latex].

a. The conjugate acid of [latex]\mathrm{CN}^{-}[/latex]is [latex]\mathrm{HCN}[/latex], whose [latex]K_{a}[/latex] is [latex]4.9 \times 10^{-10}[/latex] (Table 17.1). Hence, [latex]K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0 \times 10^{-14}}{4.9 \times 10^{-10}}=\mathbf{2 . 0} \times \mathbf{1 0}^{-5}[/latex] Note that [latex]K_{b}[/latex] is approximately equal to [latex]K_{b}[/latex] for ammonia [latex]\left(1.8 \times 10^{-5}\right)[/latex]. This means that the base strength of [latex]\mathrm{CN}^{-}[/latex]is comparable to that of [latex]\mathrm{NH}_{3}[/latex]. b. The conjugate base of [latex]\mathrm{NH}_{4}^{+}[/latex]is [latex]\mathrm{NH}_{3}[/latex], whose [latex]K_{b}[/latex] is [latex]1.8 \times 10^{-5}[/latex] (Table 17.2). Hence, [latex]K_{a}=\frac{K_{w}}{K_{b}}=\frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}}=\mathbf{5 . 6} \times \mathbf{1 0} 0^{-\mathbf{1 0}}[/latex] [latex]\mathrm{NH}_{4}{ }^{+}[/latex]is a relatively weak acid. Acetic acid, by way of comparison, has [latex]K_{a}[/latex] equal to [latex]1.7 \times 10^{-5}[/latex].

Question 17.7: Obtaining Ka from Kb or Kb from Ka Use Tables 17.1 and 17.2 ...

General Chemistry

Obtaining $K_{a}$ from $K_{b}$ or $K_{b}$ from $K_{a}$

Use Tables 17.1 and 17.2 to obtain the following at $25^{\circ} \mathrm{C}$ : a. $K_{b}$ for $\mathrm{CN}^{-}$ ; b. $K_{a}$ for $\mathrm{NH}_{4}^{+}$ .

Table 17.1
Acid-Ionization Constants at 25°C*

Substance	Formula	$K_a$
Acetic acid	$\mathrm{HC}_2 \mathrm{H}_3 \mathrm{O}_2$	$1.7 \times 10^{-5}$
Benzoic acid	$\mathrm{HC}_7 \mathrm{H}_5 \mathrm{O}_2$	$6.3 \times 10^{-5}$
Boric acid	$\mathrm{H}_3 \mathrm{BO}_3$	$5.9 \times 10^{-10}$
Carbonic acid	$\mathrm{H}_2 \mathrm{CO}_3$	$4.3 \times 10^{-7}$
	$\mathrm{HCO}_3{ }^{-}$	$4.8 \times 10^{-11}$
Cyanic acid	$\mathrm{HOCN}$	$3.5 \times 10^{-4}$
Formic acid	$\mathrm{HCHO}_2$	$1.7 \times 10^{-4}$
Hydrocyanic acìd	$\mathrm{HCN}$	$4.9 \times 10^{-10}$
Hydrofluoric acid	$\text { HF }$	$6.8 \times 10^{-4}$
Hydrogen sulfate ion	$\mathrm{HSO}_4{ }^{-}$	$1.1 \times 10^{-2}$
Hydrogen sulfide	$\mathrm{H}_2 \mathrm{~S}$	$8.9 \times 10^{-8}$
	$\mathrm{HS}^{-}$	$1.2 \times 10^{-13†}$
Hypochlorous acid	$\mathrm{HClO}$	$3.5 \times 10^{-8}$
Nitrous acid	$\mathrm{HNO}_2$	$4.5 \times 10^{-4}$
Oxalic acid	$\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4$	$5.6 \times 10^{-2}$
	$\mathrm{HC}_2 \mathrm{O}_4{ }^{-}$	$5.1 \times 10^{-5}$
Phosphoric acid	$\mathrm{H}_3 \mathrm{PO}_4$	$6.9 \times 10^{-3}$
	$\mathrm{H}_2 \mathrm{PO}_4{ }^{-}$	$6.2 \times 10^{-8}$
	$\mathrm{HPO}_4{ }^{2-}$	$4.8 \times 10^{-13}$
Phosphorous acid	$\mathrm{H}_2 \mathrm{PHO}_3$	$1.6 \times 10^{-2}$
	$\mathrm{HPHO}_3{ }^{-}$	$7 \times 10^{-7}$
Propionic acid	$\mathrm{HC}_3 \mathrm{H}_5 \mathrm{O}_2$	$1.3 \times 10^{-5}$
Pyruvic acid	$\mathrm{HC}_3 \mathrm{H}_3 \mathrm{O}_3$	$1.4 \times 10^{-4}$
Sulfurous acid	$\mathrm{H}_2 \mathrm{SO}_3$	$1.3 \times 10^{-2}$
	$\mathrm{HSO}_3{ }^{-}$	$6.3 \times 10^{-8}$
*The ionization constants for polyprotic acids are for successive ionizations. For example, for $\mathrm{H}_3 \mathrm{PO}_4$ , the equilibrium is $\mathrm{H}_3 \mathrm{PO}_4+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{H}_2 \mathrm{PO}_4{ }^{-}$ . For $\mathrm{H}_2 \mathrm{PO}_4{ }^{-}$ , the equilibrium is $\mathrm{H}_2 \mathrm{PO}_4{ }^{-}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{HPO}_4{ }^{2-}$ . $\dagger$ This value is in doubt. Some evidence suggests that it is about $10^{-19}$ . See R.J. Myers, J. Chem. Educ., 63, 687 (1986).

Table 17.2
Base-Ionization Constants at 25°C

Substance	Formula	$K_b$
Ammonia	$\mathrm{NH}_3$	$1.8 \times 10^{-5}$
Aniline	$\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$	$4.2 \times 10^{-10}$
Dimethylamine	$\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$	$5.1 \times 10^{-4}$
Ethylamine	$\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2$	$4.7 \times 10^{-4}$
Hydrazine	$\mathrm{~N}_2 \mathrm{H}_4$	$1.7 \times 10^{-6}$
Hydroxylamine	$\mathrm{NH}_2 \mathrm{OH}$	$1.1 \times 10^{-8}$
Methylamine	$\mathrm{CH}_3 \mathrm{NH}_2$	$4.4 \times 10^{-4}$
Pyridine	$\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}$	$1.4 \times 10^{-9}$
Urea	$\mathrm{NH}_2 \mathrm{CONH}_2$	$1.5 \times 10^{-14}$

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a. The conjugate acid of $\mathrm{CN}^{-}$ is $\mathrm{HCN}$ , whose $K_{a}$ is $4.9 \times 10^{-10}$ (Table 17.1). Hence,

$K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0 \times 10^{-14}}{4.9 \times 10^{-10}}=\mathbf{2 . 0} \times \mathbf{1 0}^{-5}$

Note that $K_{b}$ is approximately equal to $K_{b}$ for ammonia $\left(1.8 \times 10^{-5}\right)$ . This means that the base strength of $\mathrm{CN}^{-}$ is comparable to that of $\mathrm{NH}_{3}$ .

b. The conjugate base of $\mathrm{NH}_{4}^{+}$ is $\mathrm{NH}_{3}$ , whose $K_{b}$ is $1.8 \times 10^{-5}$ (Table 17.2). Hence,

$K_{a}=\frac{K_{w}}{K_{b}}=\frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}}=\mathbf{5 . 6} \times \mathbf{1 0} 0^{-\mathbf{1 0}}$

$\mathrm{NH}_{4}{ }^{+}$ is a relatively weak acid. Acetic acid, by way of comparison, has $K_{a}$ equal to $1.7 \times 10^{-5}$ .